Final balanced RXN in acidic solution. A current of 1.50 amp is passed for 3.00

Question

Final balanced RXN in acidic solution. A current of 1.50 amp is passed for 3.00 hours through an electrolysis cell containing a molten gold salt. AuCl_2. How many grams of gold are deposited at the cathode? g Au =____ Fe^+3 (aq) + Mn^+2 (aq) + 4H_2 O(I) is to be used in a voltaic cell. MnO^-4 (aq) + 8H^+ (aq) + 5e^-1 a Mn^+2 (aq) + 4H_2 O(I) = +1.51 V Fe^+3 (aq) + e^-1 a Fe^+2 (aq) + 0.77 V a. What will be the EMF (E_cell degree) of the cell under standard conditions? (everything 1.00M) E_cell degree = b. What will be E_cell at pH 1.000, with [Fe^+2] = [MnO_4^-1] = 0.500M and [Fe^+3] = [Mn^+2] = 0.100? E_cell = F. Calculate the Ksp of Hg_2 CI_2 at 298K using the SRP values given below. Show your setup! Hg_2 CI)2 (s) + 2e^-1 a 2Hg(I) + 2 CI^-1(aq) E degree = +0.28V Hg_2^+2 (aq) + 2e^-1 a 2Hg(I) E degree = +0.85V

Explanation / Answer

D.According to Faraday's law W=(ECt)/96500

Where

W=deposited amount of gold=?

E=equivalent mass of gold=197/2=98.5

C= current=1.50 amp

T= time =3 hr*60min/hr *60 s/ min =10800s

Plug the values we get W=16.54 g

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